Question 67

If $$\cos \theta = 4x and \sin \theta = \frac{4}{x}$$(x ≠ 0), then the value of $$\left(x^2 + \frac{1}{x^2}\right)$$ is:

Solution

Given, $$\cos\theta=4x$$ and $$\sin\theta=\frac{\ 4}{x}$$

We know that, $$\cos^2\theta+\sin^2\theta=1$$

$$=$$> $$\left(4x\right)^2+\left(\frac{\ 4}{x}\right)^2=1$$

$$=$$> $$16x^2+\frac{\ 16}{x^2}=1$$

$$=$$> $$16\left(x^2+\frac{\ 1}{x^2}\right)=1$$

$$=$$> $$x^{2\ }+\frac{1}{x^2}=\frac{1}{16}$$

Hence, the correct answer is Option B

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